The average surface area of a reservoir in the month of June is 20 . In the same month, the average rate of inflow is 10 /s, outflow rate is 15 /s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million . The evaporation (in cm) in that month is

Let 'x' cm evaporation takes place in month of June. Total inflow Total outflow As total outflow is more than total inflow, therefore depression in storage takes place. Depression in storage

GATE CE · Engineering Hydrology

Evaporation Transpiration And Stream Flow Measurement – GATE CE Engineering HydrologyPractice Questions & PYQs

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7 questions · 7 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2023NAT

The cross-section of a small river is sub-divided into seven segments of width $1.5 \mathrm{~m}$ each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is ____ $\mathrm{m}^{3} / \mathrm{s}$ (round off to one decimal place).

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📖 Explanation

Average width for 1 st land last sections,

\begin{aligned} \bar{\omega}_{1}=\bar{\omega}_{\mathrm{n}} & =\frac{\left(\omega_{1}+\frac{\omega_{2}}{2}\right)^{2}}{2 \omega_{1}} \\ & =\frac{\left(1.5+\frac{1.5}{2}\right)^{2}}{2 \times 1.5} \\ & =1.6875 \mathrm{~m} \end{aligned}

\begin{array}{|c|c|c|c|} \hline \begin{array}{l}\text { Avg. width } \\ \text { (m) } \\ \text { (1) }\end{array} & \begin{array}{l}\text { Avg. Depth } \\ \text { (m) } \\ \text { (2) }\end{array} & \begin{array}{l}\text { Avg. Velocity } \\ \text { (m/s) } \\ \text { (3) }\end{array} & \begin{array}{c}\text { Segmental } \\ \text { discharge } \\ \left(\mathrm{m}^{3} / \mathrm{s}\right)=(1) \times (2) \times (3)\end{array} \\ \hline 1.6875 & 0.40 & 0.40 & 0.270 \\ \hline 1.5 & 0.70 & \frac{0.76+0.70}{2}=0.73 & 0.767 \\ \hline 1.5 & 1.20 & \frac{1.19+1.13}{2}=1.16 & 2.088 \\ \hline 1.5 & 1.40 & \frac{1.25+1.10}{2}=1.175 & 2.467 \\ \hline 1.5 & 1.10 & \frac{1.13+1.09}{2}=1.11 & 1.831 \\ \hline 1.5 & 0.80 & \frac{0.69+0.65}{2}=0.67 & 0.804 \\ \hline 1.6875 & 0.45 & 0.42 & 0.319 \\ \hline & & & \Sigma=8.546 \\ \hline \end{array}

Q2GATE 2022MCQ

In a certain month, the reference crop evapotranspiration at a location is 6 mm/day. If the crop coefficient and soil coefficient are 1.2 and 0.8, respectively, the actual evapotranspiration in mm/day is

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📖 Explanation

Actual evapotranspiration $(ET_C)$ $=K_S \times K_C \times$ Reference evapotranspiration $(ET_0)$ $=0.8 \times 1.2 \times 6=5.76mm$

Q3GATE 2015MCQ

The average surface area of a reservoir in the month of June is 20 $km^{2}$ . In the same month, the average rate of inflow is 10 $m^{3}$ /s, outflow rate is 15 $m^{3}$ /s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million $m^{3}$ . The evaporation (in cm) in that month is

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📖 Explanation

Let 'x' cm evaporation takes place in month of June. Total inflow

\begin{array}{l} =\left(\frac{10 \times 30 \times 24 \times 60 \times 60}{20 \times 10^{6}} \times 100\right)+10 \\ =139.6 \mathrm{cm} \end{array}

Total outflow

\begin{array}{l} =\left(\frac{15 \times 30 \times 24 \times 60 \times 60}{20 \times 10^{6}} \times 100\right)+1.8+x\\ =196.2+x \mathrm{cm} \end{array}

As total outflow is more than total inflow, therefore depression in storage takes place. Depression in storage

\begin{aligned} =-\frac{16 \times 10^{6}}{20 \times 10^{6}} \times 100&=-80 \mathrm{cm} \\ \Rightarrow 139.6-(196.2+x)&=-80 \\ -x &=-80+56.6\\ \therefore \quad x&=23.4 \mathrm{cm} \end{aligned}

Q4GATE 2012MCQ

The ratio of actual evapo-transpiration to potential evapo-transpiration is in the range

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Q5GATE 2012MCQ

The top width and the depth of flow in a triangular channel were measured as 4 m and 1 m, respectively. The measured velocities on the centre line at the water surface, 0.2 m and 0.8 m below the surface are 0.7 m/s, 0.6 m/s and 0.4 m/s, respectively. Using two-point method of velocity measurement, the discharge (in $m^3$ /s) in the channel is

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📖 Explanation

\begin{aligned} V_{0.2}&=0.6m/s \\ V_{0.8}&=0.4 m/s \\ V&=\frac{V_{0.2}+V_{0.8}}{2} \\ &=\frac{0.6+0.4}{2}=0.5 m/s\\ Q&=AV \\ &= \frac{1}{2}\times 1\times 4\times 0.5=1.0m^{3}/s. \end{aligned}

Q6GATE 2003MCQ

The plan area of a reservoir is 1 $km^{2}$ . The water level in the reservoir is observed to decline by 20 cm in certain period. During this period the reservoir receives a surface inflow of 10 hectare-meters, and 20 hectare-meters are abstracted from the reservoir for irrigation and power. The plan evaporation and rainflow recorded during the same period at a nearby meteorological station are 12 cm and 3 cm respectively. The calibrated pan factor is 0.7. The seepage loss from the reservoir during the this period in hectare-meters is

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📖 Explanation

Inflow to reservoir, I= 10 ha-m Outflow from reservoir, Q=20 ha-m Evaporation loss, $E=1\times 10^{6}\times \frac{12}{100}\times 0.7 =8.4 ha-m$ Rainfall, $P=1\times 10^{6}\times \frac{3}{100} =3 ha-m$ Change in storage, $\Delta S=1\times 10^{6}\times \frac{20}{100} =-20 ha-m$ We know that, $(I+P)-(E+Q+\text{Seepage})=\Delta S$ $\Rightarrow \; (10+3)-(8.4+20+\text{Seepage})=-20$ $\Rightarrow \; 13-28.4-\text{Seepage} = -20$ $\Rightarrow \; \text{Seepage} =4.6 ha-m$

Q7GATE 2001MCQ

Isopleths are lines on a map through points having equal depth of

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